Undergrad Kramer's Equation at max Energy for photons is giving Intensity = 0

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Kramer's Equation indicates that the intensity of photons at maximum energy (Em) is zero, which raises questions about the implications of this result. The maximum energy of bremsstrahlung photons corresponds to the energy of the incident electron and is equal to the voltage difference in kilovolts peak (kVp). While lower energy photons have a non-zero intensity, the exact maximum energy has no practical significance in physics. The equation serves as an approximation and does not account for thermal effects or other factors that can affect the energy spectrum. This highlights the limitations of the equation in accurately describing photon intensity at maximum energy.
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Kramer's Equation at max Energy for photons is giving an Intensity is ZERO, how is this possible?
IE = KZ(Em – E)

where IE is the intensity of photons with energy E, Z is the atomic number of the target, Em is the maximum photon energy, and K is a constant. As pointed out earlier, the maximum possible energy that a bremsstrahlung photon can have is equal to the energy of the incident electron. The maximum energy in kiloelectron volts (keV) is numerically equal to the voltage difference between the anode and the cathode in kilovolts peak (kVp)). However, the intensity of such photons is zero as predicted by the previous equation, that is, IE is ZERO when E = Em.

Kramer's Equation at max Energy for photons is giving Intensity is ZERO, how is this possible?
 
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Every energy below the maximum has a non-zero probability. The chance to have the exact mathematical maximum has no relevance in physics, the formula is just an approximation anyway that e.g. completely ignores thermal effects or other influences that smear out the energy spectrum a bit.
 
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